I need help with this calculus problem on circles!?

Solving this takes a couple of steps. First, note that the center will be equidistant from the two points. If you take two points in the xy-plane, what's the set of points equidistant from them? It would be the line that's the perpendicular bisector of the segment joining the two lines. You can easily find the midpoint of the segment joining A=(8,-5) and B=(-1,4), which is (3.5, -0.5) and its slope m, which is -9/9 = -1. Then, the equation of the perpendicular line is just (y + 0.5) / (x - 3.5) = -1/m = 1

y + 0.5 = x - 3.5

y = x - 4

Let's call this line L.

Another way to look at L is to note that segment AB is a chord, and the perpendicular bisector of a chord passes through the center of the circle.

Another way to look at that line would be calculate the distance from a point on the line to the two points. Actually, let's just use the square of the distances: If (x,y) is the center, then

(x-8)^2 + (y+5)^2 = (x+1)^2 + (y-4)^2

Expand these and the x^2 and y^2 terms will drop out, leaving you with the equation of a line. It should be the same as the line in the prior step.

To find the center of the circle, just solve for the intersection of line L and the line you were given, 2x+3y=3.

Once you have the center (h,k), just calculate the radius using the center and one of the points:

r^2 = (h+1)^2 + (k-4)^2.

It's New Year's Day and I just got up, so pardon me for not filling in all the details. I'm sure you won't have any trouble doing that yourself while watching some football.

Here's a picture: http://i276.photobucket.com/albums/kk2/freond1/cir...

The lines were graphed by Excel, but I just drew in an approximate circle using Paint, so don't try to read any coordinates off the imprecise circle.

Have a great 2012,

Freond

Find eq of line joining A(8,-5) and B(-1,4) slopeAB = (-5-4)/(8+1) = -9/9 = -1

The perpendicular bisector of the chord AB passes through the centre Pt of AB is

(8+-1)/2.(-5+4)/2 Mid point is C( 7/2, -1/2) If O isthe centre find the eq of CO Slope of CO is the negative reciprocal of slope of AB Slope of CO is 1

y-(-1/2) = 1(x-7/2) y+1/2 = x-7/2 2y+1 = 2x -7 2y =2x -8 y = x-4

Now find where CO intersects the given line 2x+3y = = 3 Sub y = x-4 in given line

2x + 3(x-4) = 3 2x +3x -12 =3 5x = 15 x = 3 y = x-4 y = 3-4 y = -1

Centre Ois the point ( 3,-1) Find radius Length AO = sq root[ (8-3)^2+ ( -5-(-1))^2]

AO = sq root 25 + 16 AO = root 41 Radius squared = 41

eq of circle ( x-3)^2 + (y +1)^2 = 41